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enyata [817]
3 years ago
9

A very slippery ice cube slides in a vertical plane around the inside of a smooth, 20-cm-diameter horizontal pipe. The ice cube’

s speed at the bottom of the circle is 3.0 m/s. What is the ice cube’s speed at the top?
Physics
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

Explanation:

Given that,

Diameter of pipe

d = 20cm

Then, radius =d/2 =20/2

r = 10cm =0.1m

The speed at the bottom is

Vi = 3m/s

Speed at the top Vf?

At the bottom the cube is at a height of 0m

Then, y1 = 0m

At the top the cube is at a height which is the same as the diameter of the pipe

y2 = 0.2m

Now, let us consider, the energy conservation equation , which is the sum of kinetic energy and gravitational potential energy, given by,

K2 + U2 = K1 + U1

½m•Vf² + m•g•y2 = ½m•Vi² + m•g•y1

Divide all through by m

½•Vf² + g•y2 = ½•Vi² + g•y1

Since y1 = 0

So we have,

½•Vf² + g•y2 = ½•Vi²

½•Vf² = ½•Vi² — g•y2

Multiply through by 2

Vf² = Vi² —2g•y2

Vf = √(Vi²—2g•y2)

g is a constant =9.81m/s2

Vf = √(3²—2×9.81×0.2)

Vf = √(9—0.981)

Vf = √8.019

Vf = 2.83m/s

The speed of the ice cubes at the top of the pipe is 2.83m/s

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An elevated water tank holds 1 million gallons of water. Determine the force (in lbf) that the tower structure must be able to w
bixtya [17]

Answer:

8,345,925 lb-f

Explanation:

We know pressure P = F/A = ρgh ⇒ F = ρghA = ρgV

F = ρgV where F = force, ρ = density of water = 1000 kg/m³ , g = acceleration due to gravity = 9.8 m/s² and V = 1 × 10⁶ gallons = 1 × 10⁶ gallons × 1m³/264.2 gallons = 3785 m³

So, F = ρgV = 1000 kg/m³ × 9.8 m/s² × 3785 m³ = 37,093,000 N

We now convert Newtons to pound-force.

1 N = 0.225 lb-f

So, F = 37,093,000 N = 37,093,000 N × 0.225 lb-f/1 N = 8,345,925 lb-f

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3 years ago
A tortoise can move with a speed of 10.0cm/s, while a rabbit can move 10 times faster. In a race, both of them started at the sa
Xelga [282]

Answer:

C. 199.9 s

Explanation:

3 minutes = 3×60 = 180 seconds.

the turtle moves in that time 180×10 = 1800 cm.

in other words the rabbit gave it that much head-start (it does not matter if that was at the begin of in the middle of the race).

the rabbit moves with 10×10cm/s = 100cm/s.

the rabbit needs therefore 1800/100 = 18 seconds for the

1800 cm.

at that time the turtle has added another 18×10 = 180 cm.

for which the rabbit needs 180/100 = 1.8 seconds.

during that time the turtle has added 1.8×10 = 18 cm.

and so on.

in formal mathematics this looks like this :

1800 + 10x = 100x

after x seconds of the rabbit running both will have run the same distance, and it is a tie.

1800 = 90x

x = 20 seconds

so, at that point, the rabbit was actively running for 20 seconds and raced 20×100 = 2000 cm

and the turtle was actively running for 180 + 20 = 200 seconds, and also covered 200×10 = 2000 cm.

but our question tells us that the turtle won by 10 cm.

so, the race was over a little bit before these 200 seconds (for a tie).

this means, the rabbit could not run the last 10 cm for the tie (because the race was over and the turtle had won).

the rabbit would have needed 10/100 seconds for these 10 cm.

as speed = distance/time

we need to divide distance by speed

distance/1 / distance/time

to get time.

so,

10cm/1 / 100cm/s = 10s/100 = 1/10 s

so, we need to deduct this 1/10 s from the 200 seconds of the turtle (and also from the 20 seconds for the rabbit).

the race lasted of course the whole time the turtle was running (while the rabbit was resting, officially still participating in the race with speed 0 for 3 minutes).

and so, the race was 199.9 s long.

8 0
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